What is the derivative of (v^{-3}+6v^{-2})^3 ?

The derivative of (v^{-3}+6v^{-2})^3 is 3(1/(v^3)+6/(v^2))^2(-3/(v^4)-(12)/(v^3))

What is the first derivative of (v^{-3}+6v^{-2})^3 ?

The first derivative of (v^{-3}+6v^{-2})^3 is 3(1/(v^3)+6/(v^2))^2(-3/(v^4)-(12)/(v^3))

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derivative of

(v^{- 3} + 6 v^{- 2})^{3}

3 (\frac{1}{v ^{3}} + \frac{6}{v ^{2}})^{2} (- \frac{3}{v ^{4}} - \frac{12}{v ^{3}})

Solution steps

\frac{d}{d v} ((v^{- 3} + 6 v^{- 2})^{3})

Apply the chain rule:

3 (v^{- 3} + 6 v^{- 2})^{2} \frac{d}{d v} (v^{- 3} + 6 v^{- 2})

= 3 (v^{- 3} + 6 v^{- 2})^{2} \frac{d}{d v} (v^{- 3} + 6 v^{- 2})

\frac{d}{d v} (v^{- 3} + 6 v^{- 2}) = - \frac{3}{v ^{4}} - \frac{12}{v ^{3}}

= 3 (v^{- 3} + 6 v^{- 2})^{2} (- \frac{3}{v ^{4}} - \frac{12}{v ^{3}})

(v^{- 3} + 6 v^{- 2})^{2} = (\frac{1}{v ^{3}} + \frac{6}{v ^{2}})^{2}

= 3 (\frac{1}{v ^{3}} + \frac{6}{v ^{2}})^{2} (- \frac{3}{v ^{4}} - \frac{12}{v ^{3}})

\int (ln (3 x))^{2} d x

f (θ) = cos^{2} (θ)

x^{2} y^{^{'}} - x y = x^{4} cos (2 x), y (π) = \frac{π}{4}

\int \frac{6 x}{( 5 + 3 x ^{2} ) ^{4}} d x

x, 5 x, - x + 4

What is the derivative of (v^{-3}+6v^{-2})^3 ?
The derivative of (v^{-3}+6v^{-2})^3 is 3(1/(v^3)+6/(v^2))^2(-3/(v^4)-(12)/(v^3))
What is the first derivative of (v^{-3}+6v^{-2})^3 ?
The first derivative of (v^{-3}+6v^{-2})^3 is 3(1/(v^3)+6/(v^2))^2(-3/(v^4)-(12)/(v^3))